Answer:
0.2M
Explanation:
Step 1:
Data obtained from the question.
Volume of acid (Va) = 100mL
Molarity of the acid (Ma) = 0.026 M
Volume of base (Vb) = 13mL
Molarity of the base (Mb) =..?
Step 2:
The balanced equation for the reaction. This is given below:
HCl + KOH —> KCl + H2O
From the balanced equation above,
The mole ratio of the acid (nA) = 1
The mole ratio of the base (nB) = 1
Step 3:
Determination of the molarity of the base, KOH. This can be obtained as follow:
MaVa/MbVb = nA/nB
0.026 x100 / Mb x 13 = 1
Cross multiply to express in linear form
Mb x 13 = 0.026 x 100
Divide both side by 13
Mb = 0.026 x 100 / 13
Mb = 0.2M
Therefore, the molarity of the base, KOH is 0.2M
Answer:
0.2M
Explanation:
KOH(aq) + HCl(aq) ⇒ KCl(aq) + H2O(l)
We express the moles of analyte (HCl) and titrant based (KOH) on their molar concentration:
M1 * V1 = M2 * V2
The molarity of the solution is calculated with the following equation:
M2 = V1 x M1 / V2
Where:
V2 = valued sample volume
V1 = volume of titrant consumed (measured with the burette)
M1 = concentration of titrant solution
M2 = concentration of sample
M2 = 100mL * 0.026M / 13mL = 0.2M
HELP PLEASE ILL GIVE 25 pointsWhich of the following practices could help reduce erosion of water banks? a. buffer strips b. natural fertilizers and pesticides c. decrease in fossil fuel emissions d. all of the above Please select the best answer from the choices provided A B C D
Answer:
A. Buffer strips
Explanation:
The practice that could help reduce erosion of water banks is buffer strips.
What is erosion?Erosion is the action of surface processes that removes soil, rock or dissolved material from one location on the Earth's crust, and then transports it to another location where it is deposited.
One of the practices that could be used to reduce the effect of erosion is buffer strips.
What buffer strips do is slow and filter storm runoff while helping to hold soil in place.
Learn more on buffer strips here; https://brainly.com/question/26872640
What are 3 stages of the water cycle are
18.35 mL of an HCN solution were titrated with 35.4mL of a 0.268M NaOH solution to reach the equivalence point. What is the molarity of the HCN solution
Answer:
0.517
Explanation:
HCN + NaOH → NaCN + H2O [balanced as written]
(35.4 mL) x (0.268 M NaOH) x (1 mol HCN / 1 mol NaOH) / (18.35 mL HCN) = 0.517 M HCN
Answer: 0.517
Explanation:
A sample of helium gas at room temperature is compressed from 100 cm3 to 20 cm3. Its new pressure is now 30 cm Hg. What was the original pressure of the gas?
Answer:
6 cm Hg
Explanation:
Boyles Law: P1V1=P2V2
(100 mL)(x)=(20 mL)(30 cm Hg)
x = 6 cm Hg
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Two identical light bulbs are connected to a battery in a series circuit.
An ammeter is wired into the circuit at measures a current of the
battery to be 0.5 Amps. The two light bulbs are then wired in parallel.
The ammeter shows that the current:
Answer:
0.10 amps
Explanation:
